Question

A solid sphere of radius $R$ has a charge $Q$ distributed in its volume with a charge density $\rho=k r^{a}$, where $k$ and a are constants and $r$ is the distance from its center. If the electric field at $r =\frac{R}{2}$ is times that at $r = R$, the value of $a$ is

• A. 3
• B. 5
• C. 2
• D. 7

Answer 1

Answer: (C)

2

Answer 2

$\oint \vec{E} \cdot d \vec{A}=\frac{1}{\varepsilon_{0}} \int(\rho d v)$
$=\frac{1}{\varepsilon_{0}} \int k r^{a} \times 4 \pi r^{2} d r$
or $E \times 4 \pi R^{2}=\left(\frac{4 \pi k}{\varepsilon_{0}}\right) \frac{R^{(a+3)}}{(a+3)}$
$\therefore E_{1}=\frac{k R^{(a+1)}}{\varepsilon_{0}(a+3)}$
$E$ For $r=\frac{R}{2} \cdot E_{2}=\frac{k\left(\frac{R}{2}\right)^{a+1}}{\varepsilon_{0}(a+3)}$
Given $E_{2}=\frac{E_{1}}{8}$
or $\frac{k\left(\frac{R}{2}\right)^{a+1}}{\varepsilon_{0}(a+3)}=\frac{1}{8} \frac{k R^{(a+1)}}{\varepsilon_{0}(a+3)}$
$\therefore \frac{1}{2^{a+3}}=\frac{1}{8}$
or $a=2$.